Set 1: College Accreditation
Q1-4. An agency entrusted to accredit colleges looks at four parameters: faculty quality (F), reputation (R), placement quality (P), and infrastructure (I). The four parameters are used to arrive at an overall score, which the agency uses to give an accreditation to the colleges. In each parameter, there are five possible letter grades given, each carrying certain points: A (50 points), B (40 points), C (30 points), D (20 points), and F (0 points). The overall score for a college is the weighted sum of the points scored in the four parameters. The weights of the parameters are 0.1, 0.2, 0.3 and 0.4 in some order, but the order is not disclosed.
Accreditation is awarded based on the following scheme:
Eight colleges apply for accreditation, and receive the following grades in the four parameters (F, R, P, and I):
It is further known that in terms of overall scores:
1. High Q is better than Best Ed.
2. Best Ed is better than Cosmopolitan.
3. Education Aid is better than A-one.
1. What is the weight of the faculty quality parameter?
- 0.2
- 0.1
- 0.4
- 0.3
EXPLANATION
Option 2
Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with statement 2.
Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I.
R > I.
Now, let us move to statement 3.
Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P.
I > P.
Now, let us combine these two inferences – we get R > I > P.
Now, let us look at Statement 1.
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I.
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades in I.
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one parameter.
High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I.
So, the -1 in R has a great impact, so the -1 in F should have much lower impact.
In other words, weightage for F should be small and that for I should be high.
Let us account for R > I > P first.
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4.
F cannot be 0.3 either, so the 3rd one can also be eliminated.
Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal.
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3.
So, the weightages are as follows.
From the above table, we can see that the weight of the faculty quality parameter is 0.1.
2. How many colleges receive the accreditation of AAA? [TITA]
EXPLANATION
Answer: 3
Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with statement 2.
Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I.
R > I.
Now, let us move to statement 3.
Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P.
I > P.
Now, let us combine these two inferences – we get R > I > P.
Now, let us look at Statement 1.
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I.
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades in I.
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one parameter.
High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I.
So, the -1 in R has a great impact, so the -1 in F should have much lower impact.
In other words, weightage for F should be small and that for I should be high.
Let us account for R > I > P first.
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4.
F cannot be 0.3 either, so the 3rd one can also be eliminated.
Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal.
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3.
So, the weightages are as follows.
From the above table, we can see that the number of colleges which received AAA accredditation is 3.
3. What is the highest overall score among the eight colleges? [TITA]
EXPLANATION
Answer: 48
Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with statement 2.
Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I.
R > I.
Now, let us move to statement 3.
Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P.
I > P.
Now, let us combine these two inferences – we get R > I > P.
Now, let us look at Statement 1.
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I.
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades in I.
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one parameter.
High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I.
So, the -1 in R has a great impact, so the -1 in F should have much lower impact.
In other words, weightage for F should be small and that for I should be high.
Let us account for R > I > P first.
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4.
F cannot be 0.3 either, so the 3rd one can also be eliminated.
Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal.
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3.
So, the weightages are as follows.
From the above table, we can see that the highest overall score among the eight colleges is 48.
4. How many colleges have overall scores between 31 and 40, both inclusive?
- 2
- 1
- 3
- 0
EXPLANATION
Option D
Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with statement 2.
Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I.
R > I.
Now, let us move to statement 3.
Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P.
I > P.
Now, let us combine these two inferences – we get R > I > P.
Now, let us look at Statement 1.
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I.
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades in I.
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one parameter.
High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I.
So, the -1 in R has a great impact, so the -1 in F should have much lower impact.
In other words, weightage for F should be small and that for I should be high.
Let us account for R > I > P first.
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4.
F cannot be 0.3 either, so the 3rd one can also be eliminated.
Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal.
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3.
So, the weightages are as follows.
From the above table, we can see that the number of colleges which have overall scores between 31 and 40 both inclusive is 0.
Set 2 : Smartphones
Q1-4. There are only four brands of entry level smartphones called Azra, Bysi, Cxqi, and Dipq in a country. Details about their market share, unit selling price, and profitability (defined as the profit as a percentage of the revenue) for the year 2016 are given in the table below:
In 2017, sales volume of entry level smartphones grew by 40% as compared to that in 2016. Cxqi offered a 40% discount on its unit selling price in 2017, which resulted in a 15% increase in its market share. Each of the other three brands lost 5% market share. However, the profitability of Cxqi came down to half of its value in 2016. The unit selling prices of the other three brands and their profitability values remained the same in 2017 as they were in 2016.
1. The brand that had the highest revenue in 2016 is:
- Cxqi
- Bysi
- Azra
- Dipq
EXPLANATION
Option 3
From the above table, we can see that Azra had the highest revenue in 2016.
2. The brand that had the highest profit in 2016 is:
- Azra
- Dipq
- Bysi
- Cxqi
EXPLANATION
Option 4
From the above table, we can see that Cxqi had the highest profit in 2016.
3. The brand that had the highest profit in 2017 is:
- Azra
- Bysi
- Cxqi
- Dipq
EXPLANATION
Option 2
From the above table, we can see that Bysi had the highest profit in 2017.
4. The complete list of brands whose profits went up in 2017 from 2016 is:
- Bysi, Cxqi, Dipq
- Azra, Bysi, Cxqi
- Cxqi, Azra, Dipq
- Azra, Bysi, Dipq
EXPLANATION
Option 4
From the above table, we can see that the profits of Azra, Bysi, Dipq went up in 2017 from 2016.
Set 3 : Fun Sports Club
Q1-4. Fun Sports (FS) provides training in three sports – Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 39 students each of whom is enrolled in at least one of the three sports. The following details are known:
1. The number of students enrolled only in L is double the number of students enrolled in all the three sports.
2. There are a total of 17 students enrolled in G.
3. The number of students enrolled only in G is one less than the number of students enrolled only in L.
4. The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
5. The maximum student enrollment is in L.
6. Ten students enrolled in G are also enrolled in at least one more sport.
1. What is the minimum number of students enrolled in both G and L but not in K? [TITA]
EXPLANATION
Answer: 4
From condition 3, we get the above diagram.
Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.
From condition 1, we get the above diagram.
From condition 6, we can get the diagram.
From condition 4, we get the above diagram.
Now we know that the total = 39.
From condition 5,
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x.
Number of students in K = 4 + 5 + 9 + x = 18 + x.
We know that 23 – x > 18 + x.
x could be 0, 1 or 2.
For 6 – x to be minimum, x should be maximum i.e. x should take 2.
Therefore the minimum number of students enrolled in both G and L but not in K is 4.
2. If the numbers of students enrolled in K and L are in the ratio 19:22, then what is the number of students enrolled in L?
- 19
- 18
- 22
- 17
EXPLANATION
Option 3
From condition 3, we get the above diagram.
Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.
From condition 1, we get the above diagram.
From condition 6, we can get the diagram.
From condition 4, we get the above diagram.
Now we know that the total = 39.
From condition 5,
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x.
Number of students in K = 4 + 5 + 9 + x = 18 + x.
We know that 23 – x > 18 + x.
x could be 0, 1 or 2.
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x.
Number of students in K = 4 + 5 + 9 + x = 18 + x.
Ratio of K : L = 19 : 22. or, x = 1.
Or, number of students in L = 22.
3. Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one.After the withdrawal, how many students were enrolled in both G and K? [TITA]
EXPLANATION
Answer: 2
From condition 3, we get the above diagram.
Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.
From condition 1, we get the above diagram.
From condition 6, we can get the diagram.
From condition 4, we get the above diagram.
Now we know that the total = 39.
From condition 5,
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x.
Number of students in K = 4 + 5 + 9 + x = 18 + x.
We know that 23 – x > 18 + x.
x could be 0, 1 or 2.
We know that G ∩ K ∩ L = 0. We know that the 4 that were originally here are distributed among the three regions currently having x, 6 – x and 5.
We know that one student leaves K. So, this student should have gone to the region (G and L but not K).
Or, (G and L but not K) will now read 7 – x.
The other three should have opted out of one or the other of G and L. Let us assume m students left G, 3 – m should have left L.
Let us rejig the diagram.
Total number of students in G = 17 – m.
Total number of students in L = 20 + m – x.
20 + m – x – (17 – m) = 3 + 2m – x = 6.
2m – x = 3. x can only take values 0, 1 and 2. 2m = 3 + x.
Or, x has to be 1. m has to be 2.
Both G and K = 3 + x – m. 3 + 1 -2 = 2.
4. Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one.After the withdrawal, how many students were enrolled in both G and L?
- 5
- 8
- 7
- 6
EXPLANATION
Option 4
From condition 3, we get the above diagram.
Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.
From condition 1, we get the above diagram.
From condition 6, we can get the diagram.
From condition 4, we get the above diagram.
Now we know that the total = 39.
From condition 5,
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x.
Number of students in K = 4 + 5 + 9 + x = 18 + x.
We know that 23 – x > 18 + x.
x could be 0, 1 or 2.
The other three should have opted out of one or the other of G and L. Let us assume m students left G, 3 – m should have left L.
Let us rejig the diagram.
Total number of students in G = 17 – m.
Total number of students in L = 20 + m – x.
20 + m – x – (17 – m) = 3 + 2m – x = 6.
2m – x = 3. x can only take values 0, 1 and 2. 2m = 3 + x.
Or, x has to be 1. m has to be 2.
Both G and L = 7 – x = 6.
Set 4 : Products and Companies
Q1-4. Each of the 23 boxes in the picture below represents a product manufactured by one of the following three companies: Alfa, Bravo and Charlie. The area of a box is proportional to the revenue from the corresponding product, while its centre represents the Product popularity and Market potential scores of the product (out of 20). The shadings of some of the boxes have got erased.
The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:
The following facts are known:
1. Alfa and Bravo had the same number of products in the Blockbuster category.
2. Charlie had more products than Bravo but fewer products than Alfa in the No-hope category.
3. Each company had an equal number of products in the Promising category.
4. Charlie did not have any product in the Doubtful category, while Alfa had one product more than Bravo in this category.
5. Bravo had a higher revenue than Alfa from products in the Doubtful category.
6. Charlie had a higher revenue than Bravo from products in the Blockbuster category.
7. Bravo and Charlie had the same revenue from products in the No-hope category.
8. Alfa and Charlie had the same total revenue considering all products.
1. Considering all companies products, which product category had the highest revenue?
- No-hope
- Doubtful
- Promising
- Blockbuster
EXPLANATION
Option 4
The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:
From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3.
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3.
From condition 3, Alfa = Bravo = Charlie = 1.
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0.
Let us capture this differently and then take it from there.
From condition 5, The big free square is Bravo.
From condition 6, The big free square is Charlie.
From condition 7, Charlie has rectangle and Square.
From condition 8, Chances are the big square in Promising is Charlie’s.
It is between Blockbuster and Doubtful. Doubtful has two small squares whereas everything in Blockbuster is sizable. Therefore Blockbuster had the highest revenue.
2. Which of the following is the correct sequence of numbers of products Bravo had in No-hope, Doubtful, Promising and Blockbuster categories respectively?
- 2 , 3 , 1 , 2
- 1 , 3 , 1 , 2
- 3 , 3 , 1 , 2
- 1 , 3 , 1 , 3
EXPLANATION
Option 2
The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:
From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3.
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3.
From condition 3, Alfa = Bravo = Charlie = 1.
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0.
Let us capture this differently and then take it from there.
From condition 5, The big free square is Bravo.
From condition 6, The big free square is Charlie.
From condition 7, Charlie has rectangle and Square.
From condition 8, Chances are the big square in Promising is Charlie’s.
From the table we can see that Choice B 1,3,1,2 is the correct sequence of no. of products.
3. Which of the following statements is NOT correct?
- Alfa’s revenue from Blockbuster products was the same as Charlie’s revenue from Promising products.
- Bravo’s revenue from Blockbuster products was greater than Alfa’s revenue from Doubtful products.
- The total revenue from No-hope products was less than the total revenue from Doubtful products.
- Bravo and Charlie had the same revenues from No-hope products.
EXPLANATION
Option 2
The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:
From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3.
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3.
From condition 3, Alfa = Bravo = Charlie = 1.
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0.
Let us capture this differently and then take it from there.
From condition 5, The big free square is Bravo.
From condition 6, The big free square is Charlie.
From condition 7, Charlie has rectangle and Square.
From condition 8, Chances are the big square in Promising is Charlie’s.
A) Alfa Block Buster revenues = 9 squares. Charlie Promising Revenues = 9 square units. This is correct.
B) Bravo Blockbuster = 10 squares, Alfa doubtful = 12 squares.This is not correct.
C) No-Hope = 15 squares, Doubtful = 29 squares. This is correct.
D) Given in the question. This is correct.
4. If the smallest box on the grid is equivalent to revenue of Rs.1 crore, then what approximately was the total revenue of Bravo in Rs. crore?
- 30
- 40
- 34
- 24
EXPLANATION
Option 3
The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:
From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3.
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3.
From condition 3, Alfa = Bravo = Charlie = 1.
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0.
Let us capture this differently and then take it from there.
From condition 5, The big free square is Bravo.
From condition 6, The big free square is Charlie.
From condition 7, Charlie has rectangle and Square.
From condition 8, Chances are the big square in Promising is Charlie’s.
Bravo’s revenues are from 7 companies.
In No-hope, the revenues should be Rs. 4 Crores.
In Blockbuster, the revenues should be 4 + 6 = Rs. 10 Crores.
In Promising, the revenues should be Rs. 2 Crores.(this could be 3)
In Doubtful, the revenues should be 9 + 6 + 2 = Rs. 17 Crores.
This adds up to Rs. 33 crores.
The answer choice should be C.
Set 5 : Amusement Park Tickets
Q1-4. Each visitor to an amusement park needs to buy a ticket. Tickets can be Platinum, Gold, or Economy. Visitors are classified as Old, Middle-aged, or Young. The following facts are known about visitors and ticket sales on a particular day:
1. 140 tickets were sold.
2. The number of Middle-aged visitors was twice the number of Old visitors, while the number of Young visitors was twice the number of Middle-aged visitors.
3. Young visitors bought 38 of the 55 Economy tickets that were sold, and they bought half the total number of Platinum tickets that were sold.
4. The number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.
1. If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Platinum tickets, then which among the following could be the total number of Platinum tickets sold?
- 32
- 38
- 34
- 36
EXPLANATION
Option 1
From second constraint,
OLD Visitors = x, MID = 2x, Young = 4x.
7x = 140 or, x = 20 , 2x = 40 , 4x = 80.
Let us fill this in. and then go to constraint 3 and 4.
From constraint 3, Total Platinum = 2y, then YNG platinum should be y.
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.
OLD-PT should be equal to MID-PT, OLD-PT and MID-PT add up to y.
Total number of PT seats = 2y = 4 (20 – 2z) = 8 (10 – z). So, this number should be a multiple of 8.
The only possible answer is 32
2. If the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, then the number Middle-aged visitors buying Gold tickets was [TITA]
EXPLANATION
Answer: 0
From second constraint,
OLD Visitors = x, MID = 2x, Young = 4x.
7x = 140 or, x = 20 , 2x = 40 , 4x = 80.
Let us fill this in. and then go to constraint 3 and 4.
From constraint 3, Total Platinum = 2y, then YNG platinum should be y.
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.
OLD-GOLD > YNG-GOLD.
Z > 42 – y Or, y + z > 42
We need to find MID-GOLD. This is equal to 85 – 2y – z – (42 – y) = 43 – (y + z).
We know that y + z > 42. So, MID-GOLD has to be 0.
3. If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Economy tickets, then the number of Old visitors buying Gold tickets was [TITA]
EXPLANATION
Answer: 3
From second constraint,
OLD Visitors = x, MID = 2x, Young = 4x.
7x = 140 or, x = 20 , 2x = 40 , 4x = 80.
Let us fill this in. and then go to constraint 3 and 4.
From constraint 3, Total Platinum = 2y, then YNG platinum should be y.
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.
OLD-PT = MID-ECO. MID-ECO = 17 – z or, OLD-PT = 17 – z.
17 – z = – 20 – 2z. Or, z = 3.
4. Which of the following statements MUST be FALSE?
- The numbers of Gold and Platinum tickets bought by Young visitors were equal
- The numbers of Middle-aged and Young visitors buying Gold tickets were equal
- The numbers of Old and Middle-aged visitors buying Economy tickets were equal
- The numbers of Old and Middle-aged visitors buying Platinum tickets were equal
EXPLANATION
Option 3
From second constraint,
OLD Visitors = x, MID = 2x, Young = 4x.
7x = 140 or, x = 20 , 2x = 40 , 4x = 80.
Let us fill this in. and then go to constraint 3 and 4.
From constraint 3, Total Platinum = 2y, then YNG platinum should be y.
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.
Let us go statement-by-statement and see if we can find some easy one that HAS to be FALSE.
(A) The numbers of Gold and Platinum tickets bought by Young visitors were equal.
YNG-PT = YNG-GOLD. These two should add up to 42. Or, both should be 21. Total PT should be 42. Total GOLD should be 43.
All of these appear to be possible. Let us look at the other choices and see if we can find an easier CLEARLY FALSE statement.
Else, we will revisit this.
(B) The numbers of Middle-aged and Young visitors buying Gold tickets were equal.
MID-GOLD = YNG-GOLD. MID-GOLD should also be 42 – y. This tells us that z should be 1. This also appears prima-facie possible.
Let us look at the others also before we revisit this one and completely fill the grid.
(C) The numbers of Old and Middle-aged visitors buying Economy tickets were equal.
OLD-ECO = MID-ECO.These two should add up to 17 so these two CANNOT be equal.
So, statement C is DEFINITELY FALSE.
(D) The numbers of Old and Middle-aged visitors buying Platinum tickets were equal.
Set 6 : Job Interview
Q1-4. Seven candidates, Akil, Balaram, Chitra, Divya, Erina, Fatima, and Ganeshan, were invited to interview for a position. Candidates were required to reach the venue before 8 am. Immediately upon arrival, they were sent to one of three interview rooms: 101, 102, and 103. The following venue log shows the arrival times for these candidates. Some of the names have not been recorded in the log and have been marked as ‘?’.
Additionally here are some statements from the candidates:
Balaram: I was the third person to enter Room 101.
Chitra: I was the last person to enter the room I was allotted to.
Erina: I was the only person in the room I was allotted to.
Fatima: Three people including Akil were already in the room that I was allotted to when I entered it.
Ganeshan: I was one among the two candidates allotted to Room 102.
1. What best can be said about the room to which Divya was allotted?
- Either Room 101 or Room 102
- Definitely Room 103
- Definitely Room 101
- Definitely Room 102
EXPLANATION
Option 3
Look at the statements given by Erina and Ganeshan.
Think about the number of candidates in the rooms.
One room has only one candidate, one room has 2 candidates, so the third room should have 4 candidates.
So, number of candidates should be distributed as 4, 2, 1.
But which room has 2 candidates and which one has 4?
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so 101 should have had at least 3 candidates.
This implies 103 has only one candidate.
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as Fatima.
A, B and F go to room number 101. C should go to 102, E should go to 103.
Fatima is the 4th one into a room. Chitra is the last to enter her room.
So, the one at 7:45 should be entering a different room. Who is this?
A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one who goes at 7:45 am.
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the third to go into 101.
Write down the different combinations.
7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B.
Let us move on to questions.
From the above table, we can see that Divya is definitely in room 101.
2. Who else was in Room 102 when Ganeshan entered?
- No one
- Akil
- Divya
- Chitra
EXPLANATION
Option 1
Look at the statements given by Erina and Ganeshan.
Think about the number of candidates in the rooms.
One room has only one candidate, one room has 2 candidates, so the third room should have 4 candidates.
So, number of candidates should be distributed as 4, 2, 1.
But which room has 2 candidates and which one has 4?
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so 101 should have had at least 3 candidates.
This implies 103 has only one candidate.
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as Fatima.
A, B and F go to room number 101. C should go to 102, E should go to 103.
Fatima is the 4th one into a room. Chitra is the last to enter her room.
So, the one at 7:45 should be entering a different room. Who is this?
A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one who goes at 7:45 am.
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the third to go into 101.
Write down the different combinations.
7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B.
Let us move on to questions.
Ganeshan was the first to enter room 102. Therefore no one was there before Ganeshan entered room 102.
3. When did Erina reach the venue?
- 7:45 am
- 7:10 am
- 7:15 am
- 7:25 am
EXPLANATION
Option 1
Look at the statements given by Erina and Ganeshan.
Think about the number of candidates in the rooms.
One room has only one candidate, one room has 2 candidates, so the third room should have 4 candidates.
So, number of candidates should be distributed as 4, 2, 1.
But which room has 2 candidates and which one has 4?
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so 101 should have had at least 3 candidates.
This implies 103 has only one candidate.
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as Fatima.
A, B and F go to room number 101. C should go to 102, E should go to 103.
Fatima is the 4th one into a room. Chitra is the last to enter her room.
So, the one at 7:45 should be entering a different room. Who is this?
A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one who goes at 7:45 am.
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the third to go into 101.
Write down the different combinations.
7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B.
Let us move on to questions.
From the table, we can see that Erina was the last one to reach i.e. at 7:45 am
4. If Ganeshan entered the venue before Divya, when did Balaram enter the venue?
- 7:25 am
- 7:45 am
- 7:10 am
- 7:15 am
EXPLANATION
Option 1
Look at the statements given by Erina and Ganeshan.
Think about the number of candidates in the rooms.
One room has only one candidate, one room has 2 candidates, so the third room should have 4 candidates.
So, number of candidates should be distributed as 4, 2, 1.
But which room has 2 candidates and which one has 4?
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so 101 should have had at least 3 candidates.
This implies 103 has only one candidate.
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as Fatima.
A, B and F go to room number 101. C should go to 102, E should go to 103.
Fatima is the 4th one into a room. Chitra is the last to enter her room.
So, the one at 7:45 should be entering a different room. Who is this?
A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one who goes at 7:45 am.
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the third to go into 101.
Write down the different combinations.
7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B.
Let us move on to questions.
If Ganeshan entered the venue before Divya, we are looking at possibility 3 from the table.
Therefore Balaram entered the venue at 7:25 am
Set 7 : Letter Codes
Q1-4. According to a coding scheme the sentence,
Peacock is designated as the national bird of India is coded as 5688999 35 1135556678 56 458 13666689 1334 79 13366
This coding scheme has the following rules:
1. The scheme is case-insensitive (does not distinguish between upper case and lower case letters).
2. Each letter has a unique code which is a single digit from among 1,2,3,……,9.
3. The digit 9 codes two letters, and every other digit codes three letters.
4. The code for a word is constructed by arranging the digits corresponding to its letters in a non-decreasing sequence.
Answer these questions on the basis of this information
1. CAT Previous year paper – CAT Exam DI LRWhat best can be concluded about the code for the letter L?
- 6
- 8
- 1 or 8
- 1
EXPLANATION
Option 4
Digits 1 to 8 are mapped to 3 letters each, 9 is mapped to only 2.
(3 × 8) + 2 = 26 which is the total number of letters in English.
Each letter is mapped to a unique code, but remember that each number is mapped to 2/3 letters.
It is best to start from the small words. Let us attack ‘is’ and ‘as’ first.
IS = 35, AS = 56. S = 5 , I = 3 , A = 6.
Now, THE is 458. OF = 79. PEACOCK does not have a ‘7’ in in. It has an O but no F, so F should be 7 and O should be 9.
THE has 458, DESIGNATED has T and E but no H. Is there a number in THE but not in DESIGNATED?
H should be 4. T and E should be 5 and 8 in some order.
Now let us look at INDIA. INDIA = 13366. I = 3 and A = 6. So, N and D should be 1 and 6 in some order. BIRD has a D, BIRD = 1334.
What does this mean? This tells us D = 1 and N = 6.
Let us have another look at T and E. T and E and 5 and 8 in some order. Is there a word that has only one of T & E but not both?
NATIONAL has T but not E. NATIONAL has 8 but not 5. BINGO! We know T has to be 8 and E has to be 5.
Now let us go word by word. PEACOCK = P56C9CK. So, PCCK should be 8899. 9 is allotted only to two numbers.
We know that O = 9. So, what can we say about 9?
If C takes 8, then both P and K becomes 9 which is not possible as 9 can be assigned to only two letters.
Therefore C has to take 9 and P , K takes the value 8.
DESIGNATED = 1553G66851. The missing number should be G. Or, G = 7.
NATIONAL = 6683966L. The missing number should be L. Or, L should be 1.
BIRD = B3R1. 3 and 4 are missing. B and R should be 3 and 4 in some order.
B and R occur only once each in this sequence, so there is no way of resolving this.
From the above inferences, we can see that L takes 1.
2. What best can be concluded about the code for the letter B?
- 1 or 3 or 4
- 1
- 3
- 3 or 4
EXPLANATION
Option 4
Digits 1 to 8 are mapped to 3 letters each, 9 is mapped to only 2.
(3 × 8) + 2 = 26 which is the total number of letters in English.
Each letter is mapped to a unique code, but remember that each number is mapped to 2/3 letters.
It is best to start from the small words. Let us attack ‘is’ and ‘as’ first.
IS = 35, AS = 56. S = 5 , I = 3 , A = 6.
Now, THE is 458. OF = 79. PEACOCK does not have a ‘7’ in in. It has an O but no F, so F should be 7 and O should be 9.
THE has 458, DESIGNATED has T and E but no H. Is there a number in THE but not in DESIGNATED?
H should be 4. T and E should be 5 and 8 in some order.
Now let us look at INDIA. INDIA = 13366. I = 3 and A = 6. So, N and D should be 1 and 6 in some order. BIRD has a D, BIRD = 1334.
What does this mean? This tells us D = 1 and N = 6.
Let us have another look at T and E. T and E and 5 and 8 in some order. Is there a word that has only one of T & E but not both?
NATIONAL has T but not E. NATIONAL has 8 but not 5. BINGO! We know T has to be 8 and E has to be 5.
Now let us go word by word. PEACOCK = P56C9CK. So, PCCK should be 8899. 9 is allotted only to two numbers.
We know that O = 9. So, what can we say about 9?
If C takes 8, then both P and K becomes 9 which is not possible as 9 can be assigned to only two letters.
Therefore C has to take 9 and P , K takes the value 8.
DESIGNATED = 1553G66851. The missing number should be G. Or, G = 7.
NATIONAL = 6683966L. The missing number should be L. Or, L should be 1.
BIRD = B3R1. 3 and 4 are missing. B and R should be 3 and 4 in some order.
B and R occur only once each in this sequence, so there is no way of resolving this.
From the above inferences, we can see that B takes either 3 or 4.
3. For how many digits can the complete list of letters associated with that digit be identified?
- 2
- 1
- 0
- 3
EXPLANATION
Option 1
Digits 1 to 8 are mapped to 3 letters each, 9 is mapped to only 2.
(3 × 8) + 2 = 26 which is the total number of letters in English.
Each letter is mapped to a unique code, but remember that each number is mapped to 2/3 letters.
It is best to start from the small words. Let us attack ‘is’ and ‘as’ first.
IS = 35, AS = 56. S = 5 , I = 3 , A = 6.
Now, THE is 458. OF = 79. PEACOCK does not have a ‘7’ in in. It has an O but no F, so F should be 7 and O should be 9.
THE has 458, DESIGNATED has T and E but no H. Is there a number in THE but not in DESIGNATED?
H should be 4. T and E should be 5 and 8 in some order.
Now let us look at INDIA. INDIA = 13366. I = 3 and A = 6. So, N and D should be 1 and 6 in some order. BIRD has a D, BIRD = 1334.
What does this mean? This tells us D = 1 and N = 6.
Let us have another look at T and E. T and E and 5 and 8 in some order. Is there a word that has only one of T & E but not both?
NATIONAL has T but not E. NATIONAL has 8 but not 5. BINGO! We know T has to be 8 and E has to be 5.
Now let us go word by word. PEACOCK = P56C9CK. So, PCCK should be 8899. 9 is allotted only to two numbers.
We know that O = 9. So, what can we say about 9?
If C takes 8, then both P and K becomes 9 which is not possible as 9 can be assigned to only two letters.
Therefore C has to take 9 and P , K takes the value 8.
DESIGNATED = 1553G66851. The missing number should be G. Or, G = 7.
NATIONAL = 6683966L. The missing number should be L. Or, L should be 1.
BIRD = B3R1. 3 and 4 are missing. B and R should be 3 and 4 in some order.
B and R occur only once each in this sequence, so there is no way of resolving this.
From the above inferences, we can see that only two letters can be associated with digits for sure.
4. Which set of letters CANNOT be coded with the same digit?
- S , U , V
- X , Y , Z
- S , E , Z
- I , B , M
EXPLANATION
Option 1
Digits 1 to 8 are mapped to 3 letters each, 9 is mapped to only 2.
(3 × 8) + 2 = 26 which is the total number of letters in English.
Each letter is mapped to a unique code, but remember that each number is mapped to 2/3 letters.
It is best to start from the small words. Let us attack ‘is’ and ‘as’ first.
IS = 35, AS = 56. S = 5 , I = 3 , A = 6.
Now, THE is 458. OF = 79. PEACOCK does not have a ‘7’ in in. It has an O but no F, so F should be 7 and O should be 9.
THE has 458, DESIGNATED has T and E but no H. Is there a number in THE but not in DESIGNATED?
H should be 4. T and E should be 5 and 8 in some order.
Now let us look at INDIA. INDIA = 13366. I = 3 and A = 6. So, N and D should be 1 and 6 in some order. BIRD has a D, BIRD = 1334.
What does this mean? This tells us D = 1 and N = 6.
Let us have another look at T and E. T and E and 5 and 8 in some order. Is there a word that has only one of T & E but not both?
NATIONAL has T but not E. NATIONAL has 8 but not 5. BINGO! We know T has to be 8 and E has to be 5.
Now let us go word by word. PEACOCK = P56C9CK. So, PCCK should be 8899. 9 is allotted only to two numbers.
We know that O = 9. So, what can we say about 9?
If C takes 8, then both P and K becomes 9 which is not possible as 9 can be assigned to only two letters.
Therefore C has to take 9 and P , K takes the value 8.
DESIGNATED = 1553G66851. The missing number should be G. Or, G = 7.
NATIONAL = 6683966L. The missing number should be L. Or, L should be 1.
BIRD = B3R1. 3 and 4 are missing. B and R should be 3 and 4 in some order.
B and R occur only once each in this sequence, so there is no way of resolving this.
Consider option A S , U, V. We know that S takes 5. If both U and V takes 5 then there will be 4 letters coded to 5.
This is not possible.
Set 8 : Currency Exchange
The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.
A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly with another. The base exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding base exchange rates, and their selling exchange rates are 10% above their corresponding base exchange rates.
The following facts are known about the outlet on a particular day:
1. The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
2. The amounts of L used by the outlet to buy A and B are in the ratio 5:3.
3. The amounts of L the outlet received from the sales of A and B are in the ratio 5:9.
4. The outlet received 88000 units of L by selling A during the day.
5. The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
6. The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B,and 51000 units of C.
1. How many units of currency A did the outlet buy on that day? [TITA]
EXPLANATION
Answer: 1200
In currency transactions, getting the basic outline is very important.
So, if the base rate of A w.r.t to L is 100, then to buy one unit of A, we need 95 units of L and if we sell one unit of A, we will get 110 units of L.
Let us say the base rate of A w.r.t to L is 100k, then the base rate of B w.r.t to L is 120k and that of C is 1k.
To buy one unit of A, we need 95k units of L; if we sell one unit of A, we will get 110k units of L.
To buy one unit of B, we need 114k units of L; if we sell one unit of B, we will get 132k units of L.
To buy one unit of C, we need 0.95k units of L; if we sell one unit of C, we will get 1.1k units of L.
Let us keep this in mind and then build on this.
From statements 3 and 4,
Or,
To buy one unit of A, we need 190 units of L; if we sell one unit of A, we will get 220 units of L.
To buy one unit of B, we need 228 units of L; if we sell one unit of B, we will get 264 units of L.
To buy one unit of C, we need 1.9 units of L; if we sell one unit of C, we will get 2.2k units of L.
Now, let us think about currency C.
The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
We also know that we add 3000 units of C during the day.
Currency A: We spend 228000 units of L to buy 1200 units of A, we receive 88000 units of L by selling 400 units of A. We add a total of 800 units of A.
Currency B: We spend 136800 units of L to buy 600 units of B, we receive 158400 units of L by selling 600 units of B. We add 0 units of B.
Currency C: We spend 41800 units of L to buy 22000 units of C, we receive 41800 units of L by selling 19000 units of C. We add 3000 units of C.
Let us move on to the questions.
From the inferences, we can say that the outlet buys 1200 units of currency A on that day.
2. How many units of currency C did the outlet sell on that day?
- 3000
- 22000
- 6000
- 19000
EXPLANATION
Option 4
In currency transactions, getting the basic outline is very important.
So, if the base rate of A w.r.t to L is 100, then to buy one unit of A, we need 95 units of L and if we sell one unit of A, we will get 110 units of L.
Let us say the base rate of A w.r.t to L is 100k, then the base rate of B w.r.t to L is 120k and that of C is 1k.
To buy one unit of A, we need 95k units of L; if we sell one unit of A, we will get 110k units of L.
To buy one unit of B, we need 114k units of L; if we sell one unit of B, we will get 132k units of L.
To buy one unit of C, we need 0.95k units of L; if we sell one unit of C, we will get 1.1k units of L.
Let us keep this in mind and then build on this.
From statements 3 and 4,
Or,
To buy one unit of A, we need 190 units of L; if we sell one unit of A, we will get 220 units of L.
To buy one unit of B, we need 228 units of L; if we sell one unit of B, we will get 264 units of L.
To buy one unit of C, we need 1.9 units of L; if we sell one unit of C, we will get 2.2k units of L.
Now, let us think about currency C.
The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
We also know that we add 3000 units of C during the day.
Currency A: We spend 228000 units of L to buy 1200 units of A, we receive 88000 units of L by selling 400 units of A. We add a total of 800 units of A.
Currency B: We spend 136800 units of L to buy 600 units of B, we receive 158400 units of L by selling 600 units of B. We add 0 units of B.
Currency C: We spend 41800 units of L to buy 22000 units of C, we receive 41800 units of L by selling 19000 units of C. We add 3000 units of C.
Let us move on to the questions.
From the inferences, we can say that the outlet sells 19000 units of currency C on that day.
3. What was the base exchange rate of currency B with respect to currency L on that day? [TITA]
EXPLANATION
Answer: 240
4. What was the buying exchange rate of currency C with respect to currency L on that day?
- 1.10
- 0.95
- 2.20
- 1.90
EXPLANATION
Option 4
